\newproblem{lay:4_4_13}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 4.4.13}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	The set $B=\{1+t^2,t+t^2,1+2t+t^2\}$ is a basis for $\mathbb{P}_2$. Find the coordinate vector of $p(t)=1+4t+7t^2$ relative to $B$.
}{
  % Solution
	Consider the standard basis of $\mathbb{P}_2$ ($\{1,t,t^2\}$). The change-of-coordinate matrix from $B$ to the standard basis is the one whose columns
	are the expression of each one of the elements in the basis $B$ in the standard basis
	\begin{center}
		$P_{E\leftarrow B}=\begin{pmatrix}1 & 0 & 1\\ 0 & 1 & 2 \\ 1 & 1 & 1 \end{pmatrix}$
	\end{center}
	We use this matrix to convert $B$-coordinates into $E$-coordinates
	\begin{center}
		$[\mathbf{x}]_E=P_{E\leftarrow B}[\mathbf{x}]_B$
	\end{center}
	Conversely, we may invert this equation to find the $B$-coordinates of the polynomial $p(t)$.
	\begin{center}
		$[\mathbf{x}]_B=P_{E\leftarrow B}^{-1}[\mathbf{x}]_E=\begin{pmatrix}\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -1 & 0 & 1 \\
		   \frac{1}{2} & \frac{1}{2} & -\frac{1}{2}\end{pmatrix}\begin{pmatrix}1 \\ 4 \\ 7\end{pmatrix}=\begin{pmatrix}2 \\ 6 \\ -1\end{pmatrix}$
	\end{center}
}
\useproblem{lay:4_4_13}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
